JEE Main 2026 — Capacitance Question with Solution

Initial capacitance of the air capacitor is given by:

$C=\frac{{ϵ}_{0}A}{d}$

When the capacitor is half-filled with a dielectric of thickness $d/2$, it can be considered as two capacitors connected in series.

Capacitance of the dielectric-filled part:

$C_1=\frac{K{ϵ}_{0}A}{d/2}=\frac{2K{ϵ}_{0}A}{d}=2KC$

Capacitance of the air-filled part:

$C_2=\frac{{ϵ}_{0}A}{d/2}=\frac{2{ϵ}_{0}A}{d}=2C$

The equivalent capacitance $C_{eq}$ of the series combination is:

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{(2KC)(2C)}{2KC+2C}=\frac{2K}{K+1}C$

Substituting $K=5$:

$C_{eq}=\frac{2(5)}{5+1}C=\frac{10}{6}C=\frac{5}{3}C$

The increase in capacitance is:

$\Delta C=C_{eq}-C=\frac{5}{3}C-C=\frac{2}{3}C$

The percentage increase in capacitance is:

$\%\text{ increase}=\frac{\Delta C}{C}\times 100=\frac{2}{3}\times 100\approx 66.67\%$

Answer: $66.67$